Termination w.r.t. Q proof of TRS/TRCSREx2_Luc03b

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(X)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DirectTerminationProof

Q restricted rewrite system:
The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(X)
activate(X) → X

Q is empty.

We use [23] with the following order to prove termination.

Recursive Path Order [2].
Precedence:

[fst₂, activate₁, add₂, len_1] > 0
[fst₂, activate₁, add₂, len_1] > nil
[fst₂, activate₁, add₂, len_1] > n_{_fst₂}
[fst₂, activate₁, add₂, len_1] > from₁ > s₁
[fst₂, activate₁, add₂, len_1] > from₁ > cons₂
[fst₂, activate₁, add₂, len_1] > from₁ > n_{_from₁}
[fst₂, activate₁, add₂, len_1] > n_{_add₂}
[fst₂, activate₁, add₂, len_1] > n_{_len₁}